Q: Why must I choose a buffer with pKa close to my target pH?

Buffers work best when pH is within pKa ± 1 (the effective buffer range). This is because the ratio [A⁻]/[HA] needs to be between 0.1 and 10 (1:10 to 10:1) for both forms to exist in significant amounts—if ratio 10, one form dominates and the other is nearly absent, leaving little to neutralize added acid or base. For example, if you need pH 5, choose a buffer with pKa 4-6. Acetate (pKa 4.76) is perfect—at pH 5, ratio = 10^(5-4.76) = 1.74 (reasonable). But Tris (pKa 8.1) is terrible—at pH 5, ratio = 10^(5-8.1) = 0.00079 (99.9% acid, 0.1% base, no buffering). Matching pKa to pH ensures robust buffering.

Q: Can this calculator help me prepare real lab buffers?

This calculator is for educational, homework, and conceptual understanding only—not actual laboratory buffer preparation. It helps you learn Henderson–Hasselbalch, practice buffer pH calculations, compare buffer systems, and build chemical intuition. Real buffer preparation requires: (1) calibrated pH meters for verification, (2) analytical balances for precise masses, (3) volumetric glassware for accurate dilutions, (4) high-purity reagents, (5) consideration of ionic strength, temperature effects, and sterility (for biological use), (6) adherence to validated protocols and safety procedures. Never use calculator outputs directly in lab, clinical, food, or safety-critical applications. Use this tool to understand buffer theory—consult trained professionals and proper equipment for practical buffer work.

Q: What does the ratio [A⁻]/[HA] actually mean in Henderson–Hasselbalch?

The ratio [A⁻]/[HA] is the relative proportions of conjugate base to weak acid in your buffer, expressed as concentrations (M). It determines pH through Henderson–Hasselbalch. Ratio = 1 means equal amounts (pH = pKa). Ratio = 2 means twice as much base as acid (pH = pKa + 0.30). Ratio = 0.5 means half as much base as acid (pH = pKa - 0.30). Ratio = 10 means 10× more base (pH = pKa + 1, edge of buffer range). The ratio is what changes when you add strong acid/base: adding H⁺ converts A⁻ to HA (ratio decreases, pH drops); adding OH⁻ converts HA to A⁻ (ratio increases, pH rises). Because pH depends on ratio, not absolute concentrations, diluting a buffer doesn't change pH (ratio stays constant) but reduces capacity.

Q: Does dilution change buffer pH or capacity?

Dilution with pure water does NOT change buffer pH (ideally) because the ratio [A⁻]/[HA] remains constant—both concentrations decrease by the same factor, leaving their ratio unchanged. Since pH = pKa + log₁₀(ratio), and ratio is constant, pH stays the same. However, buffer capacity DOES decrease proportionally with dilution because capacity depends on total concentration (β ≈ 0.576 × C_total at pH = pKa). If you dilute a buffer 10×, pH remains the same but capacity drops 10×—it can absorb only 1/10 as much acid/base before pH shifts significantly. Practical lesson: to increase capacity, add more buffer salts, don't just dilute less. For experiments needing strong buffering, use concentrated buffers.

Q: How do I calculate buffer pH when I know the pKa and ratio?

Use the Henderson–Hasselbalch equation directly: pH = pKa + log₁₀([A⁻]/[HA]). Step 1: Find or calculate the ratio. If given concentrations, divide [conjugate base] / [weak acid]. If given moles, divide moles_base / moles_acid (in same volume). Step 2: Take log₁₀ (common logarithm, NOT natural log) of the ratio. Step 3: Add to pKa. Example: acetate buffer (pKa 4.76) with [acetate] = 0.15 M and [acetic acid] = 0.10 M. Ratio = 0.15/0.10 = 1.5. log₁₀(1.5) = 0.176. pH = 4.76 + 0.176 = 4.94. Check: pH > pKa because more base than acid (✓). This method works for any buffer system once you know pKa and concentrations or ratio.

Question 1

What is a buffer solution in simple terms?

Accepted Answer

A buffer solution is a mixture that resists changes in pH when small amounts of acid or base are added. It contains a weak acid (like acetic acid) and its conjugate base (like acetate ion), or a weak base and its conjugate acid. When acid (H⁺) is added, the conjugate base neutralizes it. When base (OH⁻) is added, the weak acid neutralizes it. This dual-action resistance maintains relatively stable pH, which is crucial for enzyme activity, cell function, and chemical reactions that are pH-sensitive. For example, blood is buffered at pH 7.4 by carbonate, phosphate, and protein systems—without buffering, eating acidic foods would drastically shift blood pH.

Question 2

How does the Henderson–Hasselbalch equation work?

Accepted Answer

The Henderson–Hasselbalch equation, pH = pKa + log₁₀([A⁻]/[HA]), connects three key buffer properties: pH (the acidity you want), pKa (the characteristic strength of your weak acid, which you choose), and [A⁻]/[HA] (the ratio of conjugate base to acid, which you control). Derived from the acid dissociation equilibrium by taking logarithms and rearranging, it tells you: (1) if base and acid are equal ([A⁻] = [HA]), then pH = pKa, (2) if more base than acid, pH > pKa, (3) if more acid than base, pH < pKa. The beauty is that pH depends on the ratio, not absolute amounts—doubling both concentrations doesn't change pH but increases buffer capacity. This equation is your main tool for buffer design and analysis.

Question 3

Why must I choose a buffer with pKa close to my target pH?

Accepted Answer

Buffers work best when pH is within pKa ± 1 (the effective buffer range). This is because the ratio [A⁻]/[HA] needs to be between 0.1 and 10 (1:10 to 10:1) for both forms to exist in significant amounts—if ratio < 0.1 or > 10, one form dominates and the other is nearly absent, leaving little to neutralize added acid or base. For example, if you need pH 5, choose a buffer with pKa 4-6. Acetate (pKa 4.76) is perfect—at pH 5, ratio = 10^(5-4.76) = 1.74 (reasonable). But Tris (pKa 8.1) is terrible—at pH 5, ratio = 10^(5-8.1) = 0.00079 (99.9% acid, 0.1% base, no buffering). Matching pKa to pH ensures robust buffering.

Question 4

What is the difference between pH and pKa in buffer calculations?

Accepted Answer

pH is the actual acidity of your solution (pH = -log₁₀[H⁺]), which changes based on what's in the solution. pKa is the characteristic acid strength of a specific weak acid (pKa = -log₁₀Ka), an intrinsic property that doesn't change (except with temperature). In buffers, pH is what you want to achieve, and pKa is what you're given by your choice of buffer system. The Henderson–Hasselbalch equation connects them: pH = pKa + log₁₀(ratio). If pH = pKa, the ratio is 1 (equal amounts). If pH > pKa, you need more base than acid. If pH < pKa, you need more acid than base. Think of pKa as the 'anchor point' around which your buffer works—pH can vary within pKa ± 1.

Question 5

Can this calculator help me prepare real lab buffers?

Accepted Answer

This calculator is for educational, homework, and conceptual understanding only—not actual laboratory buffer preparation. It helps you learn Henderson–Hasselbalch, practice buffer pH calculations, compare buffer systems, and build chemical intuition. Real buffer preparation requires: (1) calibrated pH meters for verification, (2) analytical balances for precise masses, (3) volumetric glassware for accurate dilutions, (4) high-purity reagents, (5) consideration of ionic strength, temperature effects, and sterility (for biological use), (6) adherence to validated protocols and safety procedures. Never use calculator outputs directly in lab, clinical, food, or safety-critical applications. Use this tool to understand buffer theory—consult trained professionals and proper equipment for practical buffer work.

Question 6

What does the ratio [A⁻]/[HA] actually mean in Henderson–Hasselbalch?

Accepted Answer

The ratio [A⁻]/[HA] is the relative proportions of conjugate base to weak acid in your buffer, expressed as concentrations (M). It determines pH through Henderson–Hasselbalch. Ratio = 1 means equal amounts (pH = pKa). Ratio = 2 means twice as much base as acid (pH = pKa + 0.30). Ratio = 0.5 means half as much base as acid (pH = pKa - 0.30). Ratio = 10 means 10× more base (pH = pKa + 1, edge of buffer range). The ratio is what changes when you add strong acid/base: adding H⁺ converts A⁻ to HA (ratio decreases, pH drops); adding OH⁻ converts HA to A⁻ (ratio increases, pH rises). Because pH depends on ratio, not absolute concentrations, diluting a buffer doesn't change pH (ratio stays constant) but reduces capacity.

Question 7

Does dilution change buffer pH or capacity?

Accepted Answer

Dilution with pure water does NOT change buffer pH (ideally) because the ratio [A⁻]/[HA] remains constant—both concentrations decrease by the same factor, leaving their ratio unchanged. Since pH = pKa + log₁₀(ratio), and ratio is constant, pH stays the same. However, buffer capacity DOES decrease proportionally with dilution because capacity depends on total concentration (β ≈ 0.576 × C_total at pH = pKa). If you dilute a buffer 10×, pH remains the same but capacity drops 10×—it can absorb only 1/10 as much acid/base before pH shifts significantly. Practical lesson: to increase capacity, add more buffer salts, don't just dilute less. For experiments needing strong buffering, use concentrated buffers.

Question 8

How do I calculate buffer pH when I know the pKa and ratio?

Accepted Answer

Use the Henderson–Hasselbalch equation directly: pH = pKa + log₁₀([A⁻]/[HA]). Step 1: Find or calculate the ratio. If given concentrations, divide [conjugate base] / [weak acid]. If given moles, divide moles_base / moles_acid (in same volume). Step 2: Take log₁₀ (common logarithm, NOT natural log) of the ratio. Step 3: Add to pKa. Example: acetate buffer (pKa 4.76) with [acetate] = 0.15 M and [acetic acid] = 0.10 M. Ratio = 0.15/0.10 = 1.5. log₁₀(1.5) = 0.176. pH = 4.76 + 0.176 = 4.94. Check: pH > pKa because more base than acid (✓). This method works for any buffer system once you know pKa and concentrations or ratio.

Question 9

What happens to buffer pH when I add strong acid or base?

Accepted Answer

Adding strong acid (H⁺) or base (OH⁻) to a buffer triggers a neutralization reaction that changes the ratio [A⁻]/[HA], causing a small pH shift (much smaller than unbuffered). Adding H⁺: it reacts with conjugate base (A⁻ + H⁺ → HA), decreasing [A⁻] and increasing [HA] by the amount of H⁺ added. Ratio decreases, so pH drops (but only slightly). Adding OH⁻: it reacts with weak acid (HA + OH⁻ → A⁻ + H₂O), decreasing [HA] and increasing [A⁻]. Ratio increases, so pH rises (but only slightly). The buffer 'absorbs' the addition through equilibrium shifts, preventing large pH swings. To calculate new pH: (1) update [A⁻] and [HA] using stoichiometry (1:1 reaction), (2) calculate new ratio, (3) apply Henderson–Hasselbalch with new ratio.

Question 10

Why can't I use a strong acid/base to make a buffer?

Accepted Answer

Buffers require a weak acid/conjugate base (or weak base/conjugate acid) pair in equilibrium. Strong acids (HCl, H₂SO₄, HNO₃) and strong bases (NaOH, KOH) dissociate completely—they don't establish equilibrium, so there's no weak acid or conjugate base to absorb pH changes. Example: mixing HCl with NaCl doesn't create a buffer; HCl dissociates fully (HCl → H⁺ + Cl⁻), and Cl⁻ is such a weak base (conjugate of strong acid) that it doesn't react with H⁺. Adding base would just neutralize all the HCl instantly, causing a huge pH jump. Only weak acids (like CH₃COOH, H₃PO₄⁻, NH₄⁺) partially dissociate, maintaining equilibrium between HA and A⁻—this equilibrium is what enables buffering via Le Chatelier shifts.

Question 11

What is buffer capacity and how does it relate to concentration?

Accepted Answer

Buffer capacity (β) quantifies how many moles of strong acid or base can be added per liter to shift pH by 1 unit. It's maximized when pH = pKa (equal [A⁻] and [HA]) and increases with total buffer concentration. Formula: β ≈ 2.303 × [HA] × [A⁻] / ([HA] + [A⁻]). At pH = pKa, this simplifies to β ≈ 0.576 × C_total. Practical meaning: a 0.1 M buffer at pH = pKa has β ≈ 0.058 mol/(L·pH)—adding 0.058 mol/L of acid/base shifts pH by 1 unit. A 0.01 M buffer (10× more dilute) has β ≈ 0.0058 (10× less capacity). Higher concentration = higher capacity = better resistance to pH changes. Biochemists use 50-100 mM buffers for experiments because high capacity ensures pH stability despite metabolic acids/bases.

Question 12

How do I find the required ratio [A⁻]/[HA] for a specific target pH?

Accepted Answer

Rearrange Henderson–Hasselbalch to solve for ratio: [A⁻]/[HA] = 10^(pH - pKa). Step 1: Subtract pKa from target pH to get (pH - pKa). Step 2: Raise 10 to this power (use 10^x button on calculator). Step 3: This is your required ratio. Example: you want pH 7.4 using phosphate buffer (pKa 7.2). pH - pKa = 7.4 - 7.2 = 0.2. Ratio = 10^0.2 ≈ 1.58. Meaning: you need 1.58× as much HPO₄²⁻ (base) as H₂PO₄⁻ (acid). For example, 0.158 M HPO₄²⁻ and 0.100 M H₂PO₄⁻. This method works for any buffer system—just ensure target pH is within pKa ± 1 for effective buffering.

Question 13

Can I use Henderson–Hasselbalch at any pH or temperature?

Accepted Answer

Henderson–Hasselbalch is an approximation that works well under typical student problem conditions: (1) pH between 3 and 11, (2) buffer concentrations 0.01-1 M, (3) dilute aqueous solutions where activities ≈ concentrations, (4) temperature near 25°C (adjust pKa if specified otherwise). It assumes: complete dissociation of salts (e.g., NaCH₃COO → Na⁺ + CH₃COO⁻), no side reactions, and that [H⁺] from water auto-ionization is negligible compared to buffer [H⁺]. At extreme pH (< 3 or > 11), very dilute buffers (< 0.001 M), or high ionic strength, assumptions break down and exact equilibrium calculations or activity corrections are needed. For homework, exams, and conceptual learning, HH is perfectly appropriate and gives accurate answers.

Question 14

What are some common buffer systems and when should I use them?

Accepted Answer

Acetate (CH₃COOH/CH₃COO⁻, pKa 4.76): pH 3.8-5.8, enzyme assays, protein electrophoresis. Phosphate (H₂PO₄⁻/HPO₄²⁻, pKa₂ 7.2): pH 6.2-8.2, versatile for near-neutral biological work, HPLC. Tris (Tris-HCl, pKa 8.06 at 25°C): pH 7-9, standard in molecular biology (DNA/RNA work, PCR, Western blots). HEPES (pKa 7.5): pH 6.8-8.2, cell culture (Good's buffer, minimal metal chelation, low toxicity). Citrate (multiprotic, pKa 3.1/4.8/6.4): pH 2-6, food chemistry, acidic extractions. Ammonia (NH₃/NH₄⁺, pKa 9.25): pH 8.25-10.25, basic pH work. Carbonate (H₂CO₃/HCO₃⁻, pKa 6.4): physiological buffering in blood. Choose based on target pH (match pKa ± 1) and application constraints (e.g., Tris for DNA, HEPES for cells).

Question 15

How does temperature affect buffer pH and pKa?

Accepted Answer

Temperature affects pKa of buffer systems through the van't Hoff equation (ΔH°/RT relationship). Most acid dissociations are endothermic, so pKa decreases with increasing temperature (acid gets stronger). Tris buffer has strong temperature dependence: dpKa/dT ≈ -0.03 per °C. If Tris pKa = 8.06 at 25°C, at 4°C it's ~8.7, and at 37°C it's ~7.7—a >1 pH unit range! Phosphate is more stable: dpKa/dT ≈ -0.003 per °C. If you prepare a buffer at 25°C and use it at 37°C without adjusting, pH will shift. Best practice: prepare and adjust buffers at their working temperature. For homework, assume 25°C unless stated otherwise. In real lab work, pH meters must be temperature-compensated, and some protocols specify 'pH at 25°C' or 'pH at 37°C.'

Question 16

Why do my homework problems give me pH and ask for ratio instead of the reverse?

Accepted Answer

This tests your ability to rearrange Henderson–Hasselbalch and understand the inverse relationship between pH/pKa and ratio. It's also how you'd approach buffer design: you know the pH you want (given by experiment requirements) and need to figure out what ratio of acid/base to mix. To solve: rearrange pH = pKa + log₁₀(ratio) to get log₁₀(ratio) = pH - pKa, then ratio = 10^(pH - pKa). This is the same equation, just solving for a different variable. Example problem: 'A buffer at pH 5.0 uses acetic acid (pKa 4.76). Find [acetate]/[acetic acid].' Answer: ratio = 10^(5.0 - 4.76) = 10^0.24 ≈ 1.74. You'd mix 1.74 times as much acetate as acetic acid. Mastering rearrangements builds algebraic fluency and prepares you for varied problem types.

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